The section, Special Applications, accounts for approximately 8 questions on the Power Engineering, Electrical PE exam. These questions can cover any of the topics below, Lightning/Surge Protection, Reliability, Illumination Engineering, Demand/Energy Management and Engineering Economics. All of these topics are expansive and an engineer can spend their entire career working solely in one of these fields. Thus to study these 5 topics for only 8 possible questions is quite difficult. This section focuses on the main concepts and key skills in each topic. Some topics also lend themselves to be tested on a multiple choice type exam because of their definite and commonly known equations. For example, it is most likely a safe assumption that you will have 1 or 2 Lighting Design problems, 1 or 2 Engineering Economics type problems and 1 or 2 Demand/Energy Management type problems. The other two topics are not easily tested and it is easier to combine these topics with other topics that have commonly known equation or the exam can ask concept level questions.
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Lightning protection is important because of the possible damage to property and harm to people. A lightning strike can cause fires in buildings, on transmission lines and at electrical power plants. In a building, a lightning strike can cause a surge in electrical current through the building’s electrical system that could damage sensitive electrical equipment. The same is also true for power plants and transmission lines. Lightning strikes cannot prevented, however lightning strikes can be redirected to a safe discharge locations.
Most people are familiar with lightning air terminal at the top of buildings, which are used to attract lightning strikes to a point, where the lightning surge can be safely dissipated. However, there are two other parts of a lighting protection system besides the air terminal. The second part is the conductor that transmits the energy to the third part which is the grounding system.
Lightning protection designs have to meet a certain set of standards depending on the authority having jurisdiction. This is a list of some of the codes and design standards that are used in the United States, but the most important code for the PE exam is the NFPA 780, Standard for the Installation of Lightning Protection Systems. If you are able to get this code, it should be helpful in furthering your understanding of the application of lightning protection systems, beyond the explanation in this book.
IEEE 142, IEEE Recommended Practice for Grounding of Industrial and Commercial Power Systems.
This grounding reference is discussed in a later section. This book is often called, IEEE Green Book.
NFPA 70, National Electrical Code.
This reference is discussed in the Codes and Standards section.
NFPA 780, Standard for the Installation of Lightning Protection Systems.
This reference is the most important reference for the topic of lightning protection on the PE exam.
UL 96, Lightning Protection Components.
This reference provides material standards for lightning protection components and should not be on the PE exam.
UL 467, Grounding and Bonding Equipment.
Similar to other UL references, this reference provides material standards for grounding and bonding equipment and should not be on the PE exam.
In practice, the first step in a lightning protection system design is to determine if a lightning protection system is required by code. Most clients do not want to spend additional money on a system if it is not absolutely required. As an engineer you must conduct the risk assessment to protect your client and yourself. The risk assessment is dependent on the building, where the building is located, occupancy and property.
Building: Is the building construction more susceptible to lightning strikes? The second factor is the building construction’s protection. An all metal exterior surface building will protect the contents because the metal will provide a path for the lightning as long as the interior and the framework of the building construction is not conductive. If the interior was conductive, then the lightning could jump from the conductive metal surface to another conductive surface.
Location: Is the location of the building in a high risk area that has shown a history of many lightning strikes?
This website shows real time lightning strikes in the world. The information gathered from satellites is used to develop the isokeraunic maps. https://www.lightningmaps.org
Occupancy: Does the building hold large amounts of people at many times?
Property: What is the replacement cost of the building and are the property contents of the building valuable?
See technical study guide for continuation on Lightning Protection, including topics like lightning protection systems and surge protection.
The study of reliability in power systems involves statistical analysis of various scenarios and probability of failures. However, for the purposes of the PE exam, this topic should not include any difficult statistical analysis, since the type of calculator and the time per problem limits the complexity. Thus the PE exam will most likely focus on general concepts of reliability.
The first concept is redundancy. This concept involves designing a power system such that there are redundant paths and equipment to critical locations. This means that in the event that there is a break in a circuit or a piece of equipment malfunctions, the break or non-functioning equipment can be isolated and power can continue to flow around this interruption. The most common and easily tested application of reliability is in the configuration of substations. Thus, you should know the following configurations and the pros and cons but just remember the underlying concept of reliability, because the reliability concept may be applied in other ways on the exam.
In the event of maintenance on CB-1A, CB-1B or CB-1C, the entire line “1” will be down. If there is a fault on the bus, then the entire bus will be down and no current will flow through either line “1” or “2”. You will notice that for each line into the bus, there is one circuit breaker (CB-1B for line 1 and CB-2B for line 2). There is also one circuit breaker for each line leaving the bus (CB-1C and CB-2C).
The next bus arrangement is called the double bus, double breaker arrangement. There are two breakers for each circuit and there are two buses. If there is a fault in one bus, then current can be fed through the other bus. If a single circuit breaker needs to be maintained, then current can be routed through the other breaker. During normal conditions, both buses are fed and used to provide current.
There is another arrangement in between double bus, double breaker and single bus, single breaker. This arrangement is called double bus, single breaker. You should be able to see how this arrangement is reliable if there is a fault in the bus, but how it isn’t reliable if a breaker needs maintenance.
In the building design field, lighting is a small but important part of an electrical engineer’s career. Electrical engineers select the correct types of lights and number of lights to meet appropriate levels of lighting (foot-candles) for a specific occupancy. For example, an office requires lighting levels of 40 foot-candles at desks and a restroom only requires 20 foot-candles. The different levels of lighting are determined by the amount of lighting required to carry out the occupancy’s task. In an office, workers need to be able to read document, where as in a restroom, people only need to be able to navigate the restroom and not read fine print. Once you have the required lighting level at the required location (desk level height), then you can back-calculate the light wattage and number of lights to meet this level of lighting, which will be discussed in the Lumen Method and the Point to Point Method.
There are additional aspects to illumination engineering, like color rendition, accents, glare, etc., but these aspects are beyond the scope of the PE exam, since these aspects are too specialized. If you would like to learn more about illumination engineering, you should read the illumination engineering society (IES) documents. IES is the standard for lighting design and their fundamentals of lighting book is a great introduction to the world of lighting design.
See IES Website for more information on the Illumination Engineering Society.
Before the PE skills are shown, you must understand the following lighting terms. These terms are common on lighting product data. If you need a sample of a light product data, please see the following website.
Luminous Flux: The luminous flux is the amount of energy emitted from a light per a unit of time in all directions. A lights product data gives luminous flux in units of lumens (lm). This is the value most commonly used in lighting calculations.
Luminous Intensity: The luminous intensity is given in units of candela (cd). This value describes the brightness of a light source. This value is not used in lighting product data. This term is used more often to describe lasers. A light can have a brightness of 100 candela and emit 620 lumens unrestricted. However, if the light is installed in a ceiling fixture such that half of the light is blocked by the ceiling, then the lumens will be cut in half to 310 lumens. However, the light still has the same intensity because the light intensity is not dependent on what happens outside of the light. The light still has an intensity of 100 cd. In lighting design, you care about the amount of light hitting a surface, thus the intensity does not matter, the lumens matter. Manufacturers provide lumens for their light bulbs.
Light Loss Factor: The amount of lumens provided by the light is the amount of lumens at the light. However, the lumens must travel from the lamp to the fixture, then leave the fixture and finally to the reference plane. The fixture is the metal/plastic container that holds the light bulb. This fixture can have reflective surfaces or colored surfaces to change the color and/or direction of the light. The light loss factor describes the losses from the ideal condition of the bulb to the actual condition that the bulb is installed. There are many losses that can occur and they are highlighted below.
Ballast Factor: If a light is installed in conjunction with ballast, then there will be a ballast factor. The ballast will consume a small amount of energy that will detract from the number of lumens provided by the light.
Ambient Temperature: This factor is not normally included in calculations. But the ambient temperature can affect the light output. Colder temperatures generally increase light output.
Voltage Variation: If the voltage is less than or greater than the normal voltage, then the light output will be affected.
Optical Factor: This is the amount of light absorbed by the light fixture.
Fixture Surface Factor: This is the factor of light that is absorbed by the surface of the fixture as opposed to the amount reflected.
Fixture Depreciation Factor: As a fixture ages, the surface becomes less reflective and tends to absorb more light.
Illuminance: This term describes the amount of luminous flux (lumens) that hits a certain area. In lighting design, this is the amount of light that illuminates an area. This is very applicable to lighting design, because it can tell you how much light hits a desk in an office or school or it can tell you how much light hits the gym floor. This light will directly relate to the visibility required for the task in the space.
Lumen Method or Zonal Cavity Method: The lumen method is a hand calculation method for predicting the performance of a general lighting system providing reasonably uniform illumination based on the shape of the room, the reflectance of the room and any efficiency losses/gains. This method calculates the average illuminance (foot-candles) for a space. The method entails four steps...
Point to Point Method: The point to point method is the second method used to calculate the amount of light required in a lighting design. This method is most commonly used for exterior lighting calculations like in a parking lot or in a walkway. In practice, these calculations are done by computer programs, but for the PE exam you should understand the basics of these calculations....
The demand is the maximum KW value over a certain time period. The demand is important because this value determines the maximum generation capacity that is required for a utility. Most often the demand value is not important for customers of electricity but can play a role for large customers.
The amount of energy used over a time period is found through a kilowatt-hour meter. This value is the amount of power used over a certain time period. This value is important because it informs utilities as to how much power is being generated over time. The utility can compare this value to the amount of fuel (renewable or non-renewable) to gain insight into the utilities generation efficiency. The amount of energy is valuable to customers, because this amount directly affects their electricity bill. Power engineers may be tasked to manage energy for customers by ensuring a high power factor, implementing energy efficiency projects like LED lighting retrofits or by designing renewable energy systems to reduce electricity bills.
On the PE exam, you should only need a brief understanding of the demand concept and how to save energy through techniques like power factor improvement and equipment efficiency changes. Luckily, the energy saving techniques was already discussed and you only need to understand the concept of demand.
The demand factor is determined to be the average load divided by the maximum load. This informs the utility as to how efficiently the utility is using its generation capacity.
Utility Company Demand Factor Perspective
As a utility company, you need to have enough capacity to match the peak load. But you also want your average load to equal your maximum load, such that your load factor is 1.0. This is very difficult to achieve because of how power is typically used. During the day, most people are working or are at school and not using power at home. When people return home, all residences will use power and will put a strain on the power system. The utility must have enough generation capacity to satisfy this spike in demand.
The best way to understand demand is to look at the load profiles for different types of usages. An industrial load profile can have a very high load factor, especially if it is operating 24 hours a day. Industrial loads typically do not have spikes in energy usage, because the machines are constantly running.
The next load profile is for residential customers. This load profile has a peak in the morning and a peak in the evening and valleys at night and around noon.
As you can see between the two examples, both load profiles require a 10 KVA generator but the residential profile only requires 168 KWH vs. the industrial load profile requires 216 KWH.
As a professional engineer, you will be tasked with determining the course of action for a design. Often times this will entail choosing one alternative instead of several other design alternatives. Engineers need to be able to present engineering economic analysis to their clients in order to justify why a certain alternative is more financially sound than other alternatives. The following sub-sections will present the engineering economic concepts that should be understood by the engineer and does not present a comprehensive look into the study of engineering economics.
Interest Rate and Time Value of Money
Before discussing interest rate, it is important that the engineer understand that money today is worth more than that same value of money in the future, due to factors such as inflation and interest. This is the time value of money concept. For example, if you were given the option to have $1,000 today or to have $1,000 ten years from now, most people will choose $1,000 today, without understand why this option is worth more. The reason $1,000 today is worth more is because of what could have done with that money; in the financial world, this means the amount of interest that could have been earned with that money. If you took $1,000 today and invested it at 4% per year, you would have $1,040 dollars at the end of the first year.
If you kept the $1,040 in the investment for another year, then you would have $1,081.60.
At the end of the 10 years the investment would have earned, $1,480.24.
An important formula to remember is the Future Value (FV) is equal to the Present Value (PV) multiplied by (1+interest rate), raised to the number of years.
As an example, what would be the present value of $1,000, 10 years from now, if the interest rate is 4%.
Thus in the previous example, receiving $1,000, 10 years from now, is only worth $675.46 today
It is important to understand present value because when analyzing alternatives, cash values will vary with time and the best way to make a uniform analysis is to first convert all values to consistent terms, like present value.
For example, if you were asked whether you would like $1,000 today or $1,500 in ten years (interest rate at 4%), then it would be a much more difficult question than the previous question. But with an understanding of present value, the "correct" answer would be to accept $1,500 ten years from now, because the $1000 today at 4% interest is only worth $1,480 ten years from now. In this example, the $1,000 today was converted to its future value 10 years from now. Once this value was converted, it was then compared to the $1,500, which was presented as future value in 10 years. Notice how all values were converted to future value for comparison.
The previous section described the difference between present value and future value. It also showed how a lump sum given at certain times are worth different amounts in present terms. In engineering, there are often times when annual sums are given in lieu of one time lump sums. An example would be annual energy savings due to the implementation of a more efficient HVAC system. Thus, it is important for the engineer to be able to determine the present/future value of future annual gains or losses.
For example, let's assume that a solar hot water project, provides an annual savings of $200. Using the equations from the previous section, each annual savings can be converted to either present or future value. Then these values can be summed up to determine the future and present value of annual savings of $200 for four years at an interest rate of 4%.
For longer terms, this method could become tedious. Luckily there is a formula that can be used to speed up the process in converting annuities (A) to present value and future value.
Reverse Equations, where annual value is solved:
Equipment Type Questions
In the Thermal & Fluids field, often times the engineer must develop an economic analysis on purchasing one piece of equipment over another. In this event the engineer will use terms like present value, annualized cost, future value, initial cost and other terms like salvage value, equipment lifetime and rate of return.
Salvage value is the amount a piece of equipment will be worth at the end of its lifetime. Lifetime is typically given by a manufacturer as the average lifespan (years) of a piece of equipment. Looking at the figure below, initial cost is shown as a downward arrow at year 0. Annual gains are shown as the upward arrow and maintenance costs and other costs to run the piece of equipment are shown as downward arrows starting at year 1 and proceeding to the end of the lifetime. Finally, at the end of the lifetime there is an upward arrow indicating the salvage value.
As previously stated, the most important thing in engineering economic analysis is to convert all monetary gains and costs to like terms, whether it is present value, future value, annual value or rate of return. Each specific conversion will be discussed in the following sections.
Each of the sections will use the same example, in order to illustrate the difference in converting between each of the different terms.
Example: A new chiller has an initial cost of $50,000 and a yearly maintenance cost of $1,000. At the end of its 15 year lifetime, the chiller will have a salvage value of $5,000. It is estimated that by installing this new chiller, there will be an energy savings of $5,000 per year. The interest rate is 4%.
Convert to Present Value
What is the Present Value (Present Worth) of this chiller? The first term, initial cost is already in present value.
The second term, maintenance cost must be converted from an annual cost to present value. However, we can add the annual energy savings to this amount to save time.
The third term, salvage value must be converted from future value to present value.
Finally, summing up all the like terms.
A negative Present Value indicates that the investment does not recoup the initial investment.
Convert to Future Value
See technical study guide for more detail.
Convert to Annual Value
See technical study guide for more detail.
Convert to Rate of Return
See technical study guide for more detail.
When conducting engineering economic analyses, factor values are used in lieu of formulas. Factor values are pre-calculated values that correspond to:
(1) A specific equation (convert present value to annual, convert present value to future, etc.)
(2) An interest rate.
(Number of years.
Looking up these values in a table is sometimes quicker than using the equations and lessens the possibility of calculator error. It is recommended that the engineer have the tables which can be found online. A summary of the factory values are shown below.